Proceedings of the 44th Symposium on Ring Theory and ...

By Example 3, <strong>the</strong> right R-module (R/ rad(R)) R equals <strong>the</strong> character module <strong>of</strong> <strong>the</strong>
left R-module R (R/ rad(R)). By Proposition 9 applied to <strong>the</strong> left R-module A = R R, we
have ( R (R/ rad(R)))̂ ∼ = soc( ̂R R ) ∼ = soc(R R ), because ̂R is assumed to be right free. We
thus have an isomorphism (R/ rad(R)) R
∼ = soc(RR ) <strong>of</strong> right R-modules. One can ei<strong>the</strong>r
repeat <strong>the</strong> argument for a left isomorphism (using (2)) or appeal to <strong>the</strong> <strong>the</strong>orem <strong>of</strong> Honold
[15, Theorem 2] mentioned after Definition 1.
Now assume (1). Referring to (2.1), we see that R being Frobenius implies that soc(R)
is a sum <strong>of</strong> matrix modules <strong>of</strong> <strong>the</strong> form M mi (F qi ). By Theorem 13 and summing, soc(R)
admits a left generating character. By Propositions 7 and 14, R itself admits a left
generating character. Thus (2) holds.
□
2.4. Frobenius Algebras. In this subsection I want to point out <strong>the</strong> similarity between
a general (not necessarily finite) Frobenius algebra and a finite Frobenius ring. I thank
Pr<strong>of</strong>essor Yun Fan for suggesting this short exposition.
Definition 16. A finite-dimensional algebra A over a field F is a Frobenius algebra if
<strong>the</strong>re exists a linear functional λ : A → F such that ker λ contains no nonzero left ideals
<strong>of</strong> A.
It is apparent that <strong>the</strong> structure functional λ plays a role for a Frobenius algebra
comparable to that played by a left generating character ϱ <strong>of</strong> a finite Frobenius ring. As
one might expect, <strong>the</strong> connection between λ and ϱ is even stronger when one considers
a finite Frobenius algebra. Recall that every finite field F q admits a generating character
ϑ q , by Example 2.
Theorem 17 (†). Let R be a Frobenius algebra over a finite field F q , with structure
functional λ : R → F q . Then R is a finite Frobenius ring with left generating character
ϱ = ϑ q ◦ λ.
Conversely, suppose R is a finite-dimensional algebra over a finite field F q and that R
is a Frobenius ring with generating character ϱ. Then R is a Frobenius algebra, and <strong>the</strong>re
exists a structure functional λ : R → F q such that ϱ = ϑ q ◦ λ.
Pro<strong>of</strong>. Both R ∗ := Hom Fq (R, F q ) and ̂R = Hom Z (R, Q/Z) are (R, R)-bimodules satisfying
|R ∗ | = | ̂R| = |R|. A generating character ϑ q <strong>of</strong> F q induces a bimodule homomorphism
f : R ∗ → ̂R via λ ↦→ ϑ q ◦ λ. We claim that f is injective. To that end, suppose λ ∈ ker f.
Then ϑ q ◦λ = 0, so that λ(R) ⊂ ker ϑ q . Note that λ(R) is an F q -vector subspace contained
in ker ϑ q ⊂ F q . Because ϑ q is a generating character <strong>of</strong> F q , λ(R) = 0, by Proposition 7.
Thus λ = 0, and f is injective. Because |R ∗ | = | ̂R|, f is in fact a bimodule isomorphism.
We next claim that <strong>the</strong> structure functionals in R ∗ correspond under f to <strong>the</strong> generating
characters in ̂R. That is, if ϖ = f(λ), where λ ∈ R ∗ and ϖ ∈ ̂R, <strong>the</strong>n λ satisfies <strong>the</strong>
condition that ker λ contains no nonzero left ideals <strong>of</strong> R if and only if ϖ is a generating
character <strong>of</strong> R (i.e., ker ϖ contains no nonzero left ideals <strong>of</strong> R).
Suppose ϖ is a generating character <strong>of</strong> R, and suppose that I is a left ideal <strong>of</strong> R with
I ⊂ ker λ. Since ϖ = ϑ q ◦ λ, we also have I ⊂ ker ϖ. Because ϖ is a generating character,
Proposition 7 implies I = 0, as desired.
Conversely, suppose λ satisfies <strong>the</strong> condition that ker λ contains no nonzero left ideals
<strong>of</strong> R, and suppose that I is a left ideal <strong>of</strong> R with I ⊂ ker ϖ. Then λ(I) is an F q -linear
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